Derive resultant force, variable mass system

A variable-mass system gains or loses mass continuously. Consider a system of mass $m$ moving with velocity $\vec v$ that absorbs particles of mass $\Delta m$ moving with absolute velocity $\vec u$ during time $\Delta t$.

 before:  m ---> v        dm ---> u
 after:   (m + dm) ---> v + dv

Impulse-momentum for the whole set of particles

Momentum at $t$: $m\vec v+\Delta m\,\vec u$. Momentum at $t+\Delta t$: $(m+\Delta m)(\vec v+\Delta\vec v)$. External resultant force $\sum\vec F$ acts for $\Delta t$:

$$m\vec v+\Delta m\,\vec u+\sum\vec F\,\Delta t=(m+\Delta m)(\vec v+\Delta\vec v)$$

Neglecting the second-order term $\Delta m\,\Delta\vec v$ and dividing by $\Delta t$:

$$\sum\vec F=m\frac{d\vec v}{dt}+\frac{dm}{dt}(\vec v-\vec u)$$

Using the velocity of the absorbed particles relative to the system, $\vec u_{rel}=\vec u-\vec v$:

$$\boxed{\sum\vec F=m\vec a-\frac{dm}{dt}\,\vec u_{rel}}$$

or $m\vec a=\sum\vec F+\dfrac{dm}{dt}\vec u_{rel}$.

• Mass gained, particles initially at rest ($\vec u=0$): $\sum\vec F=\dfrac{d(m\vec v)}{dt}$.
• Mass ejected (rocket): $dm/dt<0$ and $\vec u_{rel}$ is the exhaust velocity relative to the body (backward), so the term $\dfrac{dm}{dt}\vec u_{rel}$ is a forward thrust $T=\left|\dfrac{dm}{dt}\right|u_{rel}$. Then $m\vec a=\sum\vec F+\vec T$.